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JEE Advanced Physics Center of Mass Momentum and Collision 2024 JEE Advanced 2024 (Paper 2)

JEE Advanced Physics Question (2024) — Solution

Question

Two particles, 1 and 2 , each of mass m, are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at x_0, are oscillating with amplitude a and angular frequency . Thus, their positions at time t are given by x_1(t)= (x_0+d )+a t and x_2(t)= (x_0-d )-a t, respectively, where d>2 a. Particle 3 of mass m moves towards this system with speed u_0=a / 2, and undergoes instantaneous elastic collision with particle 2 , at time t_0. Finally, particles 1 and 2 acquire a center of mass speed v_ cm and oscillate with amplitude b and the same angular frequency . If the collision occurs at time t_0= /(2 ), then the value of 4 b^2 / a^2 will be ________ .

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

t_0= 2 = T 4 Particles are at extreme position After collision in C-frame using WET, aligned & W _ spring = K \\ & 1 2 k (2 ~b )^2- 1 2 k (2 a )^2=2 1 2 ~m ( a 4 )^2 ( k = spring constant ) \\ & 4 ~kb ^2-4 ka ^2=2 m a ^2 16 2 k m \\ & 4 ~b ^2= 17 4 a ^2 \\ & 4 ~b ^2 a ^2 =4.25 aligned

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